question_answer

A ball is projected up at an angle\[\theta \]with horizontal from the top of a tower with speed V. It hits the ground at point A after time\[{{t}_{A}}\]with speed\[{{V}_{A}}\]. Now, this ball is projected at same angle and speed from the base of the tower (located at point P) and it hits ground at point B after time\[{{t}_{B}}\]with speed\[{{V}_{B}}\]. Then

A)  \[PA=PB\]

B)  \[{{t}_{A}}={{t}_{B}}\]

C)  \[{{V}_{A}}>{{V}_{B}}\]

D)  ball A hits the ground at an angle\[(-\theta )\]with horizontal

Correct Answer: C

Solution :

 Consider, the two projectiles projected from\[O\] and P as shown in the figure. Range of both the particles \[R=\frac{{{v}^{2}}\sin 2\theta }{g}\] \[\Rightarrow \] \[OQ=PB=R\] Clearly, velocity at the points A/ and B will be same. Velocity of the ball following the trajectory OCA will increase after A due to accelerated motion under gravity. Therefore, we can say that\[{{v}_{A}}>{{v}_{B}}\]